What is the general solution for 1+2sin^2(t)=-5cos(t) ?

The general solution for 1+2sin^2(t)=-5cos(t) is t=(2pi)/3+2pin,t=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+2sin^2(t)=-5cos(t)

Solution

1 + 2 sin^{2} (t) = - 5 cos (t)

Solution

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Degrees

t = 12 0^{\circ} + 36 0^{\circ} n, t = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

1 + 2 sin^{2} (t) = - 5 cos (t)

Subtract

- 5 cos (t)

from both sides

1 + 2 sin^{2} (t) + 5 cos (t) = 0

Rewrite using trig identities

1 + 2 sin^{2} (t) + 5 cos (t)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 2 (1 - cos^{2} (t)) + 5 cos (t)

Simplify

1 + 2 (1 - cos^{2} (t)) + 5 cos (t) : 5 cos (t) - 2 cos^{2} (t) + 3

1 + 2 (1 - cos^{2} (t)) + 5 cos (t)

Expand

2 (1 - cos^{2} (t)) : 2 - 2 cos^{2} (t)

2 (1 - cos^{2} (t))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (t)

= 2 \cdot 1 - 2 cos^{2} (t)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 cos^{2} (t)

= 1 + 2 - 2 cos^{2} (t) + 5 cos (t)

Add the numbers:

1 + 2 = 3

= 5 cos (t) - 2 cos^{2} (t) + 3

= 5 cos (t) - 2 cos^{2} (t) + 3

3 - 2 cos^{2} (t) + 5 cos (t) = 0

Solve by substitution

3 - 2 cos^{2} (t) + 5 cos (t) = 0

Let:

cos (t) = u

3 - 2 u^{2} + 5 u = 0

3 - 2 u^{2} + 5 u = 0 : u = - \frac{1}{2}, u = 3

3 - 2 u^{2} + 5 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + 5 u + 3 = 0

Solve with the quadratic formula

- 2 u^{2} + 5 u + 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) \cdot 3}{2 ( - 2 )}

5^{2} - 4 (- 2) \cdot 3 = 7

5^{2} - 4 (- 2) \cdot 3

Apply rule

- (- a) = a

= 5^{2} + 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

Add the numbers:

25 + 24 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 5 + 7}{2 ( - 2 )}, u_{2} = \frac{- 5 - 7}{2 ( - 2 )}

u = \frac{- 5 + 7}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 5 + 7}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 + 7}{- 2 \cdot 2}

Add/Subtract the numbers:

- 5 + 7 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 5 - 7}{2 ( - 2 )} : 3

\frac{- 5 - 7}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 5 - 7}{- 2 \cdot 2}

Subtract the numbers:

- 5 - 7 = - 12

= \frac{- 12}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 12}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 3

Substitute back

u = cos (t)

cos (t) = - \frac{1}{2}, cos (t) = 3

cos (t) = - \frac{1}{2}, cos (t) = 3

cos (t) = - \frac{1}{2} : t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = - \frac{1}{2}

General solutions for

cos (t) = - \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

cos (t) = 3 :

No Solution

cos (t) = 3

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

t = \frac{2 π}{3} + 2 πn, t = \frac{4 π}{3} + 2 πn

Popular Examples

2sin(2x)-3-6/(sin(2x)-1)=0

2 sin (2 x) - 3 - \frac{6}{sin ( 2 x ) - 1} = 0

3sin^2(x)+11sin(x)+10=0

3 sin^{2} (x) + 11 sin (x) + 10 = 0

cos(x)(cos(x)+5)=2sin^2(x)

cos (x) (cos (x) + 5) = 2 sin^{2} (x)

tan^2(x)+3tan(x)-10=0

tan^{2} (x) + 3 tan (x) - 10 = 0

cos^2(x)-sin(x)=-1

cos^{2} (x) - sin (x) = - 1

Frequently Asked Questions (FAQ)

What is the general solution for 1+2sin^2(t)=-5cos(t) ?
The general solution for 1+2sin^2(t)=-5cos(t) is t=(2pi)/3+2pin,t=(4pi)/3+2pin